The following are equivalent definitions for a Galois extension field (also simply known as a Galois extension) 
of 
.
1. 
is the splitting field for a collection of separable
polynomials. When 
is a finite extension, then only one separable polynomial is necessary.
2. The field automorphisms of 
that fix 
do not fix any intermediate fields 
, i.e., 
.
3. Every irreducible polynomial over 
which has a root in 
factors into linear factors in 
. Also, 
must be a separable extension.
4. A field automorphism 
of the algebraic
closure 
of 
for which 
must fix 
.
That is to say that 
must be a field automorphism of 
fixing 
. Also, 
must be a separable extension.
A Galois extension has all of the above properties. For example, consider 
, the rationals adjoined by the imaginary number 
, over 
, which is a Galois extension. Note that 
contains all of the roots of 
, and is generated by them, so it is the splitting
field of 
.
Of course, there are two distinct roots in 
so it is separable. The only nontrivial automorphism fixing
is given by complex conjugation
 |
(1)
|
whose fixed field is 
. The only irreducible polynomials with rational coefficients
with roots of the form 
with 
are 
(
) and 
. Both split into linear factors over 
. Finally, the algebraic closure 
is the set of algebraic numbers in 
. Given an automorphism of the algebraic numbers that sends
to itself, it must fix 
for trivial reasons. In general, it is not so simple to verify
all of these properties, which makes their equivalence useful.
There are a couple of ways for an extension not to be a Galois extension. One is for it to not be a normal extension. For instance,
is not normal, and hence not Galois. It is missing the complex roots of 
. Its only nontrivial automorphism is defined by 
, which not only fixes 
but also the subfield 
.
Another possibility for a non-Galois extension is for it to be not separable. The field characteristic of such a field must
be finite since all polynomials are separable in characteristic zero. Moreover, all
finite fields are perfect,
i.e., all algebraic extensions are separable. Consider the field
of rational functions with coefficients in 
, infinite in size and characteristic 2 (
).
 |
(2)
|
and the extension
 |
(3)
|
For instance, 
and 
.
Then 
is the splitting field of 
, as 
in characteristic 2, and so it
is a normal extension. However, 
is not separable because 
has repeated roots in its splitting field, 
.
